# Find the surface area of the hyperbolic paraboloid above a closed disk

Find the surface area of the hyperbolic paraboloid $z = 13 + x^2 - y^2$ that lies above the closed disk $x^2 + y^2 \le 4$.

I know that the surface area is equal to: $\iint$$\sqrt{1+[f_x(x,y)]^2 + [f_y(x,y)]^2} dA over some region R. f_x(x,y) = 2x f_y(x,y) = -2y It's now equal to: \iint$$\sqrt{1+4x^2 + 4y^2}$ $dA$ for some region $R$

I'm stuck at this point. Can someone help me finish this? Thank you!

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Put $r=x^2+y^2$ and replace $dA$ by $r dr d\theta$. – coffeemath Dec 11 '12 at 11:55
@coffeemath At what certain values should I evaluate the double integral? – Marcus Dec 11 '12 at 12:12

Square roots in integrals generally make problems difficult, but here it's a clue that the problem can be simplified by making a change of variable to polar coordinates. Let $x = r \cos \theta$, $y=r\sin\theta$.
To get you started, we write the region of integration in polar coordinates. The closed disk $x^2+y^2\leq 4$ is given in polar coordinates by $\{(r,\theta)|0\leq r\leq 2, 0 \leq \theta < 2\pi\}$. This gives you very easy limits of integration.
You've already found the integrand in terms of $x$ and $y$. Use the polar coordinate transformation to express it in terms of $r$ and $\theta$, and use trigonometric identities to make the integrand very simple.
Finally, remember that whenever you make a change in variable you must multiply by the appropriate Jacobian determinant. (It's easy to forget this step. Remember that $dA=dxdy$ in our original coordinates, but $dA\neq drd\theta$ in our new coordinates - something's missing.)